# How many values of t does the particle change direction if a particle moves with acceleration #a(t)=3t^2-2t# and it's initial velocity is 0?

##### 1 Answer

The particle changes direction for one value of t:

#### Explanation:

If the acceleration of the particle is

#inta(t)dt = int(3t^2-2t)dt = t^3 - t^2 + C#

Since the initial velocity is 0 (when t=0):

#0^3 - 0^2 + C = 0#

#C = 0# So our equation simplifies to

#v(t) = t^3 - t^2#

Every point where the velocity is

#v(t) = 0#

#t^3 - t^2 = 0#

#t^2(t-1) = 0#

#t = 0 " " and " " t=1#

To check whether the particle changes directions at each of these points, we need to pick test points to check the intervals between them (we don't have to check before

#v(color(red)(1/2)) = (color(red)(1/2))^3 - (color(red)(1/2))^2 = 1/8 - 1/4 = -1/8# So in the interval from

#t=0# to#t=1# , the particle moves in the negative direction.

#v(color(blue)2) = (color(blue)2)^3 - (color(blue)2)^2 = 8-4 = 4# So in the interval beyond

#t=1# , the particle moves in the positive direction.

Therefore, the particle changes direction at

*Final Answer*